The current drawn from the battery in the giv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given circuit can be redrawn as a Wheatstone bridge. The resistors are arranged such that the bridge is balanced because the ratio of resistan

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) The given circuit can be redrawn as a Wheatstone bridge. The resistors are arranged such that the bridge is balanced because the ratio of resistances in the arms is $\frac{3}{3} = \frac{2}{2} = 1$. In a balanced Wheatstone bridge,no current flows through the central $5 \ \Omega$ resistor. Thus,the circuit simplifies to two parallel branches,each consisting of two resistors in series. The upper branch has a resistance of $3 \ \Omega + 2 \ \Omega = 5 \ \Omega$. The lower branch has a resistance of $3 \ \Omega + 2 \ \Omega = 5 \ \Omega$. The equivalent resistance $R_{eq}$ of these two parallel branches is given by $\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$,which gives $R_{eq} = 2.5 \ \Omega$. The current $I$ drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{6 \ V}{2.5 \ \Omega} = 2.4 \ A$.

The current drawn from the battery in the given network is (Internal resistance of battery is neglected) (in $A$)

Similar Questions

The figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion $PQ$ is

In the given figure: $V_1=V, V_2=\alpha V, R_1=\beta R, R_2=\gamma R$,where $\alpha, \beta$,and $\gamma$ are positive real numbers. The value of current $I$ is

For the network shown in the figure,the value of the current $i$ is:

In the given circuit,the values of $I_1, I_2, I_3$ are respectively:

Twelve equal resistors,each of resistance $R$,are connected to form a skeleton cube. The equivalent resistance between two diagonally opposite corners is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module